Modulus of infinite product of complex functions We know that for complex (entire) functions $f,g$ we have  $|f(z)g(z)|=|f(z)||g(z)|$, where $|.|$ means complex modulus.
What about if we have an infinite product? Is it true that
$$\bigg| \prod_{k=1}^{\infty} f_{k}(z)\bigg|= \prod_{k=1}^{\infty} |f_{k}(z)| $$
where $\{f_{k}\}$ is any set of entire functions.
 A: The formula is incorrect in general  because the right hand side might be defined while the left hand side is not.     
For example if $f_n(z)=(-1)^n$, then $\Pi f_n(z)$ is not defined and so $|\Pi f_n(z)|$ isn't either; but obviously $\Pi |f_n(z)|$ is defined and its value is $1$.  
But under auspicious circumstances I cannot exclude that something could be salvaged...  
Warning
In contrast to what happens for series, there is no naïve notion of absolute convergence for infinite products of complex numbers: else the example above shows  you would have the disastrous terminology that some absolutely convergent products  are divergent!
The best substitute is that the convergence of $\Pi (1+|a_n|) $implies the convergence of $\Pi (1+a_n) $.
Beware that some books' tratment of infinite products is not quite satisfactory. If you want to take the safe way,  I cannot recommend warmly enough 
Remmert's Classical Topics in Complex Function Theory where the theme is handled right at the beginning of the book.
